\(\int x^{5/2} (2-b x)^{5/2} \, dx\) [563]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F(-1)]
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 16, antiderivative size = 150 \[ \int x^{5/2} (2-b x)^{5/2} \, dx=-\frac {5 \sqrt {x} \sqrt {2-b x}}{16 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{48 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2-b x}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {5 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{8 b^{7/2}} \]

[Out]

1/6*x^(7/2)*(-b*x+2)^(3/2)+1/6*x^(7/2)*(-b*x+2)^(5/2)+5/8*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(7/2)-5/48*x^(
3/2)*(-b*x+2)^(1/2)/b^2-1/24*x^(5/2)*(-b*x+2)^(1/2)/b+1/8*x^(7/2)*(-b*x+2)^(1/2)-5/16*x^(1/2)*(-b*x+2)^(1/2)/b
^3

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {52, 56, 222} \[ \int x^{5/2} (2-b x)^{5/2} \, dx=\frac {5 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{8 b^{7/2}}-\frac {5 \sqrt {x} \sqrt {2-b x}}{16 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{48 b^2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{8} x^{7/2} \sqrt {2-b x}-\frac {x^{5/2} \sqrt {2-b x}}{24 b} \]

[In]

Int[x^(5/2)*(2 - b*x)^(5/2),x]

[Out]

(-5*Sqrt[x]*Sqrt[2 - b*x])/(16*b^3) - (5*x^(3/2)*Sqrt[2 - b*x])/(48*b^2) - (x^(5/2)*Sqrt[2 - b*x])/(24*b) + (x
^(7/2)*Sqrt[2 - b*x])/8 + (x^(7/2)*(2 - b*x)^(3/2))/6 + (x^(7/2)*(2 - b*x)^(5/2))/6 + (5*ArcSin[(Sqrt[b]*Sqrt[
x])/Sqrt[2]])/(8*b^(7/2))

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 56

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
 a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {5}{6} \int x^{5/2} (2-b x)^{3/2} \, dx \\ & = \frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {1}{2} \int x^{5/2} \sqrt {2-b x} \, dx \\ & = \frac {1}{8} x^{7/2} \sqrt {2-b x}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {1}{8} \int \frac {x^{5/2}}{\sqrt {2-b x}} \, dx \\ & = -\frac {x^{5/2} \sqrt {2-b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2-b x}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {5 \int \frac {x^{3/2}}{\sqrt {2-b x}} \, dx}{24 b} \\ & = -\frac {5 x^{3/2} \sqrt {2-b x}}{48 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2-b x}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {5 \int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx}{16 b^2} \\ & = -\frac {5 \sqrt {x} \sqrt {2-b x}}{16 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{48 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2-b x}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {5 \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx}{16 b^3} \\ & = -\frac {5 \sqrt {x} \sqrt {2-b x}}{16 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{48 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2-b x}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {5 \text {Subst}\left (\int \frac {1}{\sqrt {2-b x^2}} \, dx,x,\sqrt {x}\right )}{8 b^3} \\ & = -\frac {5 \sqrt {x} \sqrt {2-b x}}{16 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{48 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2-b x}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {5 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{8 b^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.68 \[ \int x^{5/2} (2-b x)^{5/2} \, dx=\frac {\sqrt {x} \sqrt {2-b x} \left (-15-5 b x-2 b^2 x^2+54 b^3 x^3-40 b^4 x^4+8 b^5 x^5\right )}{48 b^3}-\frac {5 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2-b x}}\right )}{4 b^{7/2}} \]

[In]

Integrate[x^(5/2)*(2 - b*x)^(5/2),x]

[Out]

(Sqrt[x]*Sqrt[2 - b*x]*(-15 - 5*b*x - 2*b^2*x^2 + 54*b^3*x^3 - 40*b^4*x^4 + 8*b^5*x^5))/(48*b^3) - (5*ArcTan[(
Sqrt[b]*Sqrt[x])/(Sqrt[2] - Sqrt[2 - b*x])])/(4*b^(7/2))

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.70

method result size
meijerg \(\frac {\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (-b \right )^{\frac {7}{2}} \left (-56 b^{5} x^{5}+280 b^{4} x^{4}-378 b^{3} x^{3}+14 b^{2} x^{2}+35 b x +105\right ) \sqrt {-\frac {b x}{2}+1}}{336 b^{3}}-\frac {5 \sqrt {\pi }\, \left (-b \right )^{\frac {7}{2}} \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{8 b^{\frac {7}{2}}}}{\left (-b \right )^{\frac {5}{2}} \sqrt {\pi }\, b}\) \(105\)
risch \(-\frac {\left (8 b^{5} x^{5}-40 b^{4} x^{4}+54 b^{3} x^{3}-2 b^{2} x^{2}-5 b x -15\right ) \sqrt {x}\, \left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{48 b^{3} \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {5 \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{16 b^{\frac {7}{2}} \sqrt {x}\, \sqrt {-b x +2}}\) \(131\)
default \(-\frac {x^{\frac {5}{2}} \left (-b x +2\right )^{\frac {7}{2}}}{6 b}+\frac {-\frac {x^{\frac {3}{2}} \left (-b x +2\right )^{\frac {7}{2}}}{6 b}+\frac {5 \left (-\frac {3 \sqrt {x}\, \left (-b x +2\right )^{\frac {7}{2}}}{20 b}+\frac {3 \left (\frac {\left (-b x +2\right )^{\frac {5}{2}} \sqrt {x}}{3}+\frac {5 \left (-b x +2\right )^{\frac {3}{2}} \sqrt {x}}{6}+\frac {5 \sqrt {x}\, \sqrt {-b x +2}}{2}+\frac {5 \sqrt {\left (-b x +2\right ) x}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right )}{2 \sqrt {-b x +2}\, \sqrt {x}\, \sqrt {b}}\right )}{20 b}\right )}{6 b}}{b}\) \(157\)

[In]

int(x^(5/2)*(-b*x+2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

120/(-b)^(5/2)/Pi^(1/2)/b*(1/40320*Pi^(1/2)*x^(1/2)*2^(1/2)*(-b)^(7/2)*(-56*b^5*x^5+280*b^4*x^4-378*b^3*x^3+14
*b^2*x^2+35*b*x+105)/b^3*(-1/2*b*x+1)^(1/2)-1/192*Pi^(1/2)*(-b)^(7/2)/b^(7/2)*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/
2)))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.15 \[ \int x^{5/2} (2-b x)^{5/2} \, dx=\left [\frac {{\left (8 \, b^{6} x^{5} - 40 \, b^{5} x^{4} + 54 \, b^{4} x^{3} - 2 \, b^{3} x^{2} - 5 \, b^{2} x - 15 \, b\right )} \sqrt {-b x + 2} \sqrt {x} - 15 \, \sqrt {-b} \log \left (-b x + \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right )}{48 \, b^{4}}, \frac {{\left (8 \, b^{6} x^{5} - 40 \, b^{5} x^{4} + 54 \, b^{4} x^{3} - 2 \, b^{3} x^{2} - 5 \, b^{2} x - 15 \, b\right )} \sqrt {-b x + 2} \sqrt {x} - 30 \, \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{48 \, b^{4}}\right ] \]

[In]

integrate(x^(5/2)*(-b*x+2)^(5/2),x, algorithm="fricas")

[Out]

[1/48*((8*b^6*x^5 - 40*b^5*x^4 + 54*b^4*x^3 - 2*b^3*x^2 - 5*b^2*x - 15*b)*sqrt(-b*x + 2)*sqrt(x) - 15*sqrt(-b)
*log(-b*x + sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1))/b^4, 1/48*((8*b^6*x^5 - 40*b^5*x^4 + 54*b^4*x^3 - 2*b^3*x^2
- 5*b^2*x - 15*b)*sqrt(-b*x + 2)*sqrt(x) - 30*sqrt(b)*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))))/b^4]

Sympy [F(-1)]

Timed out. \[ \int x^{5/2} (2-b x)^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(x**(5/2)*(-b*x+2)**(5/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.39 \[ \int x^{5/2} (2-b x)^{5/2} \, dx=\frac {\frac {15 \, \sqrt {-b x + 2} b^{5}}{\sqrt {x}} + \frac {85 \, {\left (-b x + 2\right )}^{\frac {3}{2}} b^{4}}{x^{\frac {3}{2}}} + \frac {198 \, {\left (-b x + 2\right )}^{\frac {5}{2}} b^{3}}{x^{\frac {5}{2}}} - \frac {198 \, {\left (-b x + 2\right )}^{\frac {7}{2}} b^{2}}{x^{\frac {7}{2}}} - \frac {85 \, {\left (-b x + 2\right )}^{\frac {9}{2}} b}{x^{\frac {9}{2}}} - \frac {15 \, {\left (-b x + 2\right )}^{\frac {11}{2}}}{x^{\frac {11}{2}}}}{24 \, {\left (b^{9} - \frac {6 \, {\left (b x - 2\right )} b^{8}}{x} + \frac {15 \, {\left (b x - 2\right )}^{2} b^{7}}{x^{2}} - \frac {20 \, {\left (b x - 2\right )}^{3} b^{6}}{x^{3}} + \frac {15 \, {\left (b x - 2\right )}^{4} b^{5}}{x^{4}} - \frac {6 \, {\left (b x - 2\right )}^{5} b^{4}}{x^{5}} + \frac {{\left (b x - 2\right )}^{6} b^{3}}{x^{6}}\right )}} - \frac {5 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{8 \, b^{\frac {7}{2}}} \]

[In]

integrate(x^(5/2)*(-b*x+2)^(5/2),x, algorithm="maxima")

[Out]

1/24*(15*sqrt(-b*x + 2)*b^5/sqrt(x) + 85*(-b*x + 2)^(3/2)*b^4/x^(3/2) + 198*(-b*x + 2)^(5/2)*b^3/x^(5/2) - 198
*(-b*x + 2)^(7/2)*b^2/x^(7/2) - 85*(-b*x + 2)^(9/2)*b/x^(9/2) - 15*(-b*x + 2)^(11/2)/x^(11/2))/(b^9 - 6*(b*x -
 2)*b^8/x + 15*(b*x - 2)^2*b^7/x^2 - 20*(b*x - 2)^3*b^6/x^3 + 15*(b*x - 2)^4*b^5/x^4 - 6*(b*x - 2)^5*b^4/x^5 +
 (b*x - 2)^6*b^3/x^6) - 5/8*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x)))/b^(7/2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 472 vs. \(2 (105) = 210\).

Time = 22.27 (sec) , antiderivative size = 472, normalized size of antiderivative = 3.15 \[ \int x^{5/2} (2-b x)^{5/2} \, dx=\frac {{\left ({\left ({\left (2 \, {\left ({\left (b x - 2\right )} {\left (4 \, {\left (b x - 2\right )} {\left (\frac {5 \, {\left (b x - 2\right )}}{b^{5}} + \frac {61}{b^{5}}\right )} + \frac {1251}{b^{5}}\right )} + \frac {3481}{b^{5}}\right )} {\left (b x - 2\right )} + \frac {11395}{b^{5}}\right )} {\left (b x - 2\right )} + \frac {11895}{b^{5}}\right )} \sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2} - \frac {6930 \, \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b} b^{4}}\right )} b {\left | b \right |} - 36 \, {\left ({\left ({\left (2 \, {\left (b x - 2\right )} {\left ({\left (b x - 2\right )} {\left (\frac {4 \, {\left (b x - 2\right )}}{b^{4}} + \frac {41}{b^{4}}\right )} + \frac {171}{b^{4}}\right )} + \frac {745}{b^{4}}\right )} {\left (b x - 2\right )} + \frac {965}{b^{4}}\right )} \sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2} - \frac {630 \, \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b} b^{3}}\right )} {\left | b \right |} + \frac {120 \, {\left ({\left ({\left (b x - 2\right )} {\left (2 \, {\left (b x - 2\right )} {\left (\frac {3 \, {\left (b x - 2\right )}}{b^{3}} + \frac {25}{b^{3}}\right )} + \frac {163}{b^{3}}\right )} + \frac {279}{b^{3}}\right )} \sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2} - \frac {210 \, \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b} b^{2}}\right )} {\left | b \right |}}{b} - \frac {320 \, {\left (\sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2} {\left ({\left (b x - 2\right )} {\left (\frac {2 \, {\left (b x - 2\right )}}{b^{2}} + \frac {13}{b^{2}}\right )} + \frac {33}{b^{2}}\right )} - \frac {30 \, \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b} b}\right )} {\left | b \right |}}{b^{2}}}{240 \, b} \]

[In]

integrate(x^(5/2)*(-b*x+2)^(5/2),x, algorithm="giac")

[Out]

1/240*((((2*((b*x - 2)*(4*(b*x - 2)*(5*(b*x - 2)/b^5 + 61/b^5) + 1251/b^5) + 3481/b^5)*(b*x - 2) + 11395/b^5)*
(b*x - 2) + 11895/b^5)*sqrt((b*x - 2)*b + 2*b)*sqrt(-b*x + 2) - 6930*log(abs(-sqrt(-b*x + 2)*sqrt(-b) + sqrt((
b*x - 2)*b + 2*b)))/(sqrt(-b)*b^4))*b*abs(b) - 36*(((2*(b*x - 2)*((b*x - 2)*(4*(b*x - 2)/b^4 + 41/b^4) + 171/b
^4) + 745/b^4)*(b*x - 2) + 965/b^4)*sqrt((b*x - 2)*b + 2*b)*sqrt(-b*x + 2) - 630*log(abs(-sqrt(-b*x + 2)*sqrt(
-b) + sqrt((b*x - 2)*b + 2*b)))/(sqrt(-b)*b^3))*abs(b) + 120*(((b*x - 2)*(2*(b*x - 2)*(3*(b*x - 2)/b^3 + 25/b^
3) + 163/b^3) + 279/b^3)*sqrt((b*x - 2)*b + 2*b)*sqrt(-b*x + 2) - 210*log(abs(-sqrt(-b*x + 2)*sqrt(-b) + sqrt(
(b*x - 2)*b + 2*b)))/(sqrt(-b)*b^2))*abs(b)/b - 320*(sqrt((b*x - 2)*b + 2*b)*sqrt(-b*x + 2)*((b*x - 2)*(2*(b*x
 - 2)/b^2 + 13/b^2) + 33/b^2) - 30*log(abs(-sqrt(-b*x + 2)*sqrt(-b) + sqrt((b*x - 2)*b + 2*b)))/(sqrt(-b)*b))*
abs(b)/b^2)/b

Mupad [F(-1)]

Timed out. \[ \int x^{5/2} (2-b x)^{5/2} \, dx=\int x^{5/2}\,{\left (2-b\,x\right )}^{5/2} \,d x \]

[In]

int(x^(5/2)*(2 - b*x)^(5/2),x)

[Out]

int(x^(5/2)*(2 - b*x)^(5/2), x)